Given, Planck's constant, h = 6.6 $$\times$$ 10^$$-$$34 Js

Boltzmann constant, k_B = 1.38 $$\times$$ 10^$$-$$23 J/K

Mass of an electron, m_e = 9 $$\times$$ 10^$$-$$31 kg

Temperature of an ideal gas, T = 300 K

As we know that, de-Broglie wavelength,

$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$ .... (i)

Here, E is the kinetic energy,

$$E = {{3{K_B}T} \over 2}$$

Substituting value of E in Eq. (i), we get

$$\lambda = {h \over {\sqrt {3m{K_B}T} }}$$

Substituting the given values in the above equation, we get

$$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$$

= 6.26 nm

$$\therefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.